CS 61C: Great Ideas in Computer
Architecture (Machine Structures)
Lecture 28: Single-Cycle CPU
Datapath Control Part 1
Instructor: Senior Lecturer SOE Dan Garcia
http://inst.eecs.Berkeley.edu/~cs61c/
Review
• CPU design involves Datapath, Control
– 5 Stages for MIPS Instructions
1.
2.
3.
4.
5.
Instruction Fetch
Instruction Decode & Register Read
ALU (Execute)
Memory
Register Write
• Datapath timing: single long clock cycle or one
short clock cycle per stage
www.technologyreview.com/news/512776/microchip-adapts-to-severe-damage/
Technology In the News
Microchip works after damage by routing around it
Researchers at CalTech have
developed an integrated circuit
(IC) that can continue to perform
in the face of structural damage.
“It doesn’t physically repair flaws,
it uses a second processor to
come up with new ways to
perform a task in spite of the
damage. It can also be
programmed to prioritize energy
savings or speed”. In some
sense, like the Internet, that can
“route around” trouble spots.
Datapath and Control
+4
imm
opcode, funct
Controller
ALU
Data
memory
rd
rs
rt
registers
PC
instruction
memory
• Datapath based on data transfers required to perform
instructions
• Controller causes the right transfers to happen
CPU Clocking (1/2)
• For each instruction, how do we control the flow of
information though the datapath?
• Single Cycle CPU: All stages of an instruction
completed within one long clock cycle
– Clock cycle sufficiently long to allow each instruction to
complete all stages without interruption within one cycle
1. Instruction
Fetch
2. Decode/
Register
Read
3. Execute 4. Memory
5. Reg.
Write
CPU Clocking (2/2)
• Alternative multiple-cycle CPU: only one stage of instruction
per clock cycle
– Clock is made as long as the slowest stage
1. Instruction 2. Decode/
Fetch
Register
Read
3. Execute
4. Memory
5. Register
Write
– Several significant advantages over single cycle execution:
Unused stages in a particular instruction can be skipped
OR instructions can be pipelined (overlapped)
Agenda
• Stages of the Datapath
• Datapath Instruction Walkthroughs
• Datapath Design
Five Components of a Computer
Computer
Processor
Control
Datapath
Memory
(passive)
(where
programs,
data live
when
running)
Devices
Input
Output
Keyboard,
Mouse
Disk
(where
programs,
data live
when not
running)
Display,
Printer
Processor Design: 5 steps
Step 1: Analyze instruction set to determine datapath
requirements
– Meaning of each instruction is given by register transfers
– Datapath must include storage element for ISA registers
– Datapath must support each register transfer
Step 2: Select set of datapath components & establish
clock methodology
Step 3: Assemble datapath components that meet the
requirements
Step 4: Analyze implementation of each instruction to
determine setting of control points that realizes the
register transfer
Step 5: Assemble the control logic
The MIPS Instruction Formats
• All MIPS instructions are 32 bits long. 3 formats:
– R-type
31
26
op
rs
6 bits
31
– I-type
26
op
31
16
rt
5 bits
5 bits
21
rs
6 bits
– J-type
21
5 bits
11
6
rd
shamt
funct
5 bits
5 bits
6 bits
16
0
address/immediate
rt
5 bits
16 bits
26
op
6 bits
0
target address
26 bits
• The different fields are:
–
–
–
–
–
–
0
op: operation (“opcode”) of the instruction
rs, rt, rd: the source and destination register specifiers
shamt: shift amount
funct: selects the variant of the operation in the “op” field
address / immediate: address offset or immediate value
target address: target address of jump instruction
The MIPS-lite Subset
• ADDU and SUBU
31
op
– addu rd,rs,rt
– subu rd,rs,rt
• OR Immediate:
26
rs
6 bits
31
op
31
– lw rt,rs,imm16
– sw rt,rs,imm16
• BRANCH:
31
26
op
– beq rs,rt,imm16 6 bits
rs
5 bits
shamt
funct
5 bits
5 bits
6 bits
0
16 bits
0
immediate
5 bits
21
0
rd
16
rt
5 bits
6
immediate
5 bits
21
rs
11
16
rt
5 bits
26
6 bits
5 bits
21
rs
op
16
rt
5 bits
26
– ori rt,rs,imm16 6 bits
• LOAD and
STORE Word
21
16 bits
16
rt
5 bits
0
immediate
16 bits
Register Transfer Language (RTL)
RTL gives the meaning of the instructions
All start by fetching the instruction
{op , rs , rt , rd , shamt , funct}  MEM[ PC ]
{op , rs , rt ,
Imm16}  MEM[ PC ]
Inst
Register Transfers
ADDU
R[rd]  R[rs] + R[rt]; PC  PC + 4
SUBU
R[rd]  R[rs] – R[rt]; PC  PC + 4
ORI
R[rt]  R[rs] | zero_ext(Imm16); PC  PC + 4
LOAD
R[rt]  MEM[ R[rs] + sign_ext(Imm16)]; PC  PC + 4
STORE
MEM[ R[rs] + sign_ext(Imm16) ]  R[rt]; PC  PC + 4
BEQ
if ( R[rs] == R[rt] )
then PC  PC + 4 + (sign_ext(Imm16) || 00)
else PC  PC + 4
Step 1: Requirements of the
Instruction Set
• Memory (MEM)
– Instructions & data (will use one for each)
• Registers (R: 32 x 32)
– Read RS
– Read RT
– Write RT or RD
• PC
• Extender (sign/zero extend)
• Add/Sub/OR unit for operation on register(s) or extended
immediate
• Add 4 (+ maybe extended immediate) to PC
• Compare registers?
Step 2: Components of the
Datapath
• Combinational Elements
• Storage Elements + Clocking Methodology
• Building Blocks
OP
CarryIn
A
A
CarryOut
32
Adder
B
32
32
Y
B
32
Multiplexer
32
ALU
32
Sum
A
MUX
Adder
B
32
Select
32
ALU
32
Result
ALU Needs for MIPS-lite + Rest of MIPS
• Addition, subtraction, logical OR, ==:
ADDU
SUBU
ORI
R[rd] = R[rs] + R[rt]; ...
R[rd] = R[rs] – R[rt]; ...
R[rt] = R[rs] | zero_ext(Imm16)...
BEQ
if ( R[rs] == R[rt] )...
• Test to see if output == 0 for any ALU
operation gives == test. How?
• P&H also adds AND, Set Less Than (1 if A < B, 0
otherwise)
• ALU follows Chapter 5
Storage Element: Idealized Memory
Write Enable
Address
• Memory (idealized)
– One input bus: Data In
– One output bus: Data Out
• Memory word is found by:
Data In
32
Clk
DataOut
32
– Address selects the word to put on Data Out
– Write Enable = 1: address selects the memory
word to be written via the Data In bus
• Clock input (CLK)
– CLK input is a factor ONLY during write operation
– During read operation, behaves as a combinational logic
block: Address valid  Data Out valid after “access time”
Storage Element: Register (Building Block)
Write Enable
• Similar to D Flip Flop except
– N-bit input and output
– Write Enable input
• Write Enable:
Data In
Data Out
N
N
clk
– Negated (or deasserted) (0): Data Out will not
change
– Asserted (1): Data Out will become Data In on
positive edge of clock
Storage Element: Register File
RW RA RB
Write Enable 5 5 5
• Register File consists of 32 registers:
– Two 32-bit output busses:
busA and busB
– One 32-bit input bus: busW
• Register is selected by:
busW
32
Clk
32 x 32-bit
Registers
busA
32
busB
32
– RA (number) selects the register to put on busA (data)
– RB (number) selects the register to put on busB (data)
– RW (number) selects the register to be written
via busW (data) when Write Enable is 1
• Clock input (clk)
– Clk input is a factor ONLY during write operation
– During read operation, behaves as a combinational logic block:
• RA or RB valid  busA or busB valid after “access time.”
Step 3a: Instruction Fetch Unit
• Register Transfer
Requirements 
Datapath Assembly
• Instruction Fetch
• Read Operands and Execute
Operation
• Common RTL operations clk
– Fetch the Instruction:
mem[PC]
– Update the program counter:
• Sequential Code:
PC  PC + 4
• Branch and Jump:
PC  “something else”
PC
Next Address
Logic
Address
Instruction Word
Instruction
Memory
32
Step 3b: Add & Subtract
• R[rd] = R[rs] op R[rt] (addu rd,rs,rt)
– Ra, Rb, and Rw come from instruction’s Rs, Rt, and Rd fields
31
26
op
6 bits
21
rs
5 bits
16
rt
5 bits
11
rd
5 bits
6
shamt
5 bits
0
funct
6 bits
– ALUctr and RegWr: control logic after decoding the instruction
Rd Rs Rt
RegWr 5 5 5
clk
Rw Ra Rb
32 x 32-bit
Registers
busA
32
busB
32
• … Already defined the register file & ALU
ALU
busW
32
ALUctr
Result
32
Clocking Methodology
Clk
.
.
.
.
.
.
.
.
.
.
.
.
• Storage elements clocked by same edge
• Flip-flops (FFs) and combinational logic have some delays
– Gates: delay from input change to output change
– Signals at FF D input must be stable before active clock edge to allow
signal to travel within the FF (set-up time), and we have the usual
clock-to-Q delay
• “Critical path” (longest path through logic) determines length
of clock period
Register-Register Timing:
One Complete Cycle
Clk
PC Old Value
Rs, Rt, Rd,
Op, Func
Old Value
ALUctr
Old Value
RegWr
Old Value
busA, B
Old Value
busW
Old Value
New Value
Instruction Memory Access Time
New Value
Delay through Control Logic
New Value
New Value
Register File Access Time
New Value
ALU Delay
New Value
ALUctr
RegWr Rd Rs Rt
5
Rw
busW
5
Ra Rb
busA
32
ALU
RegFile
clk
5
busB
32
32
Register Write
Occurs Here
Putting it All Together:A Single Cycle Datapath
RegDst
32
Equal
0
5
5
5
Rw Ra Rb
RegFile
busA
busB
32
16
Extender
imm16
MemtoReg
MemWr
Rs Rt
clk
clk
ALUctr
32
=
ALU
busW
PC
PC Ext
Adder
Mux
00
RegWr
Adder
4
Rt Rd Imm16
Rd Rt
1
Instruction<31:0>
<0:15>
nPC_sel
Rs
<11:15>
Adr
<16:20>
<21:25>
Inst
Memory
0
32
1
32
Data In
clk
imm16
ExtOp
ALUSrc
32
0
WrEn Adr
Data
Memory
1
Processor Design: 3 of 5 steps
Step 1: Analyze instruction set to determine datapath
requirements
– Meaning of each instruction is given by register transfers
– Datapath must include storage element for ISA registers
– Datapath must support each register transfer
Step 2: Select set of datapath components & establish
clock methodology
Step 3: Assemble datapath components that meet the
requirements
Step 4: Analyze implementation of each instruction to
determine setting of control points that realizes the
register transfer
Step 5: Assemble the control logic